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1 answer:
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Answer:
1) 2.467 N
2) a) 0.248m
b) 2.3π rad/sec
Explanation:
Given data:
mass of Banana bunch ( m ) = 2.0 kg
radius of circular path ( R ) = 0.5 m
number of revolutions completed = 2
Time to complete 2 revolutions = 6 seconds
1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds
F = mv^2 / r ----- ( 1 )
where V = 2πR/T
where : R = 0.5 , m = 2, T = 4 seconds
Insert values into equation 1
F = 2 * 4π^2 * 0.5/4^2
= 2.467 N
2a) Calculate the maximum distance of coin from center
angular velocity ( w ) = v/r
coefficient of static friction ( μ ) = 0.25
---- ( 1 )
mv^2/r = μmg --- ( 2 ) cancelling the mass on both sides eqn 2 becomes
v^2 = μ*g*r
dividing both sides of equation by r^2
w^2 = μ*g/r
hence determine distance ( r ) of coin from center
r = 0.25 * 9.81 / π^2 = 0.248 m
2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping
distance coin is placed ( r ) = 4.7 cm = 0.047 m
find speed of rotation ( w )
w^2 = μ*g/r
w = √ 0.25 * 9.81/ 0.047
= 7.2236 rad/secs ≈ 2.3π rad/sec